24.8. THE RIESZ REPRESENTATION THEOREM 683

Theorem 24.8.2 Let X be any Banach space and let (Ω,S ,µ) be a measure space.Then for p > 1, Lp′ (Ω;X ′) is isometric to a subspace of (Lp (Ω;X))′. Also, for g ∈Lp′ (Ω;X ′),

sup|| f ||p≤1

∣∣∣∣∫Ω

g(ω)( f (ω))dµ

∣∣∣∣= ∥g∥p′ .

If p = 1 and p′ = ∞, this is still true assuming µ (Ω)< ∞.

Proof: First observe that for f ∈ Lp (Ω;X) and g ∈ Lp′ (Ω;X ′),

ω → g(ω)( f (ω))

is a function in L1 (Ω). (To obtain measurability, write f as a limit of simple functions.Holder’s inequality then yields the function is in L1 (Ω).) Define

θ : Lp′ (Ω;X ′

)→ (Lp (Ω;X))′

by

θg( f )≡∫

Ω

g(ω)( f (ω))dµ.

Holder’s inequality implies

∥θg∥ ≤ ∥g∥p′ (24.35)

and it is also clear that θ is linear. Next it is required to show ∥θg∥= ∥g∥p′ . To begin with,always assume p > 1.

This will first be verified for simple functions. Assume ∥g∥p′ ̸= 0 since if not, there isnothing to show. Let

g(ω) =m

∑i=1

c∗i XEi (ω) , g ∈ Lp′ (Ω;X ′

),c∗i ̸= 0

where 0 ̸= c∗i ∈ X ′, the Ei are disjoint. Let di ∈ X be such that ∥di∥X = 1 and

c∗i (di)≥ ∥c∗i ∥− ε

Then let

f (ω)≡ 1

∥g∥p′−1Lp′ (Ω;X ′)

m

∑i=1

di ∥c∗i ∥p′−1 XEi (ω)

Then since p′−1 = p′/p,

∫Ω

∥ f∥p dµ =1

∥g∥p′

Lp′ (Ω;X ′)

∫Ω

m

∑i=1∥c∗i ∥

p′XEi (ω)dµ =∥g∥p′

Lp′ (Ω;X ′)

∥g∥p′

Lp′ (Ω;X ′)

= 1